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In a cafe, the customer arrives at a mean rate of 2 per $\min$. Find the probability of arrival of 5 customers in 1 minute using the Poisson distribution formula.
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Given: $\lambda=2$, and $x=5$.
Using the Poisson distribution formula:
$P(X=x)=\left(e^{-\lambda} \lambda^{x}\right) / x !$
$P(X=5)=\left(e^{-2} 2^{5}\right) / 5 !$
$P(X=6)=0.036$
Answer: The probability of arrival of 5 customers per minute is $3.6 \%$.
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